An aqueous solution contains SCN^{-},Br^{-},I | vedclass

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(A) The precipitation of a salt occurs when the ionic product exceeds its solubility product constant $(K_{sp})$. For a series of salts with the same

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$I^{-}$

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(A) The precipitation of a salt occurs when the ionic product exceeds its solubility product constant $(K_{sp})$. For a series of salts with the same stoichiometry ($1:1$ ratio),the salt with the lowest $K_{sp}$ value will precipitate first upon the addition of a common ion. Comparing the given values: $K_{sp}(AgSCN) = 7.1 	imes 10^{-7}$ $K_{sp}(AgCl) = 1.2 	imes 10^{-10}$ $K_{sp}(AgBr) = 3.5 	imes 10^{-13}$ $K_{sp}(AgI) = 1.7 	imes 10^{-16}$ Since $K_{sp}(AgI)$ is the smallest,$AgI$ will precipitate first.

Salt	$K_{sp}$ at $298 K$
$AgBr$	$5.0 \times 10^{-13}$
$Zn(OH)_2$	$1.0 \times 10^{-15}$
$Hg_2Cl_2$	$1.3 \times 10^{-18}$

Similar Questions

If the solubility product $K_{sp}$ of a sparingly soluble salt $MX_2$ at $25\,^{\circ}C$ is $1.0 \times 10^{-11}$,the solubility of the salt in $\text{mol L}^{-1}$ at this temperature will be:

The correct order of solubility of the given salts in water at $298 K$ is:
Salt $K_{sp}$ at $298 K$
$AgBr$ $5.0 \times 10^{-13}$
$Zn(OH)_2$ $1.0 \times 10^{-15}$
$Hg_2Cl_2$ $1.3 \times 10^{-18}$

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$AgCl$ is soluble in:

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